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I came across this problem on my exam and even after three hours of trying I am not able to get through the problem.

I think it can be expressed as the limit of a sum but I am not sure and all my attempts to do the same have failed.

Question:

If the limit $$\lim_{n \to \infty} \left[\left(\prod_{i=1}^{n}i!\right)^{\frac{1}{n^2}} (n^{x})\right]$$ exists and is finite what are the possible values of $x$ and the corresponding values of limit?

I got something like $$e^{\ln{1\over n}\displaystyle\sum_{r=1}^n\left[1-{r-1\over n}\ln\left({r\over n}\right)\right]}$$ but I am having trouble with that r-1 over there.

Can this be expressed as a Riemann sum?

(So recently all the answers I have got seem to be skipping steps I get that all these people are professional in their own fields but can you please try to write answers for someone who has considerably low IQ than yourself, thanks for the same)

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  • $\begingroup$ Have you tried something like $y=e^{\ln y}$? Then note that the first parenthesis can be written as $1^n\cdot 2^{n-1}\cdot...\cdot(n-1)^2\cdot n$ $\endgroup$
    – Andrei
    Commented Mar 30, 2020 at 2:01
  • $\begingroup$ @Andrei Yes indeed I have but then I got something like ${1 \over n}\sum_{r=1}^n [1-{r-1\over n} ln({r\over n})] $ $\endgroup$ Commented Mar 30, 2020 at 2:08
  • $\begingroup$ That r-1 is what is bugging me. Shouldn't it be r $\endgroup$ Commented Mar 30, 2020 at 2:10
  • $\begingroup$ It will not matter in the limit $n\to\infty$. When $n$ is large, small $r$ will not count, and for large enough $r$, you have $r-1\approx r$. $\endgroup$
    – Andrei
    Commented Mar 30, 2020 at 2:44
  • $\begingroup$ @Andrei Can you justify this rigorously? $\endgroup$ Commented Mar 30, 2020 at 2:46

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Asymptotic Expansion via Riemann Sum

Compute the log of the product as a Riemann Sum $$ \begin{align} \frac1{n^2}\sum_{k=1}^n k\log(n-k+1) &=\sum_{k=1}^n\frac{k}{n}\left(\log\left(1-\frac{k}{n}+\frac1n\right)+\log(n)\right)\frac1n\tag{1a}\\ &\sim\int_0^1x\log(1-x)\,\mathrm{d}x+\frac12\log(n)\tag{1b}\\ &=\int_0^1\log(1-x)\,\mathrm{d}\frac{x^2-1}2+\frac12\log(n)\tag{1c}\\ &=-\int_0^1\frac{x+1}2\,\mathrm{d}x+\frac12\log(n)\tag{1d}\\ &=\frac12\log(n)-\frac34\tag{1e} \end{align} $$ Thus, the product is asymptotically $$ \left(\prod_{k=1}^nk!\right)^{1/n^2}\sim e^{-3/4}n^{1/2}\tag2 $$ Therefore, for $x=-1/2$, the limit comes to $$ \bbox[5px,border:2px solid #C0A000]{\lim_{n\to\infty}\left(\prod_{k=1}^nk!\right)^{1/n^2}n^{-1/2}=e^{-3/4}}\tag3 $$ For $x\lt-1/2$, the limit is $0$.


Asymptotic Expansion via Euler-Maclaurin Sum Formula

As is shown in this answer, we have asymptotically in $n$, $$ \sum_{k=1}^n k^{-z} =\zeta(z)+\frac{n^{1-z}}{1-z}+\frac12n^{-z}-\frac{z}{12}n^{-1-z}+O\left(\frac1{n^{3+z}}\right)\tag4 $$ applying $-\frac{\mathrm{d}}{\mathrm{d}z}$: $$ \begin{align} \sum_{k=1}^n\log(k)k^{-z} &=-\zeta'(z)+n^{1-z}\frac{(1-z)\log(n)-1}{(1-z)^2}+\frac12\log(n)n^{-z}\\ &-n^{-1-z}\frac{z\log(n)-1}{12}+O\!\left(\frac{\log(n)}{n^{3+z}}\right)\tag5 \end{align} $$ Setting $z=0$: $$ \sum_{k=1}^n\log(k)=\overbrace{\,\,-\zeta'(0)\ }^{\frac12\log(2\pi)}+n(\log(n)-1)+\frac12\log(n)+\frac1{12n}+O\!\left(\frac{\log(n)}{n^3}\right)\tag6 $$ Setting $z=-1$: $$ \begin{align} \sum_{k=1}^n\log(k)k &=\overbrace{-\zeta'(-1)}^{\log(A)-\frac1{12}}+n^2\frac{2\log(n)-1}4+\frac12n\log(n)+\frac{\log(n)+1}{12}\\ &+O\!\left(\frac{\log(n)}{n^2}\right)\tag7 \end{align} $$ where $A$ is the Glaisher–Kinkelin constant.

Thus, $$ \begin{align} \sum_{k=1}^n(n-k+1)\log(k) &=n^2\frac{2\log(n)-3}4+n\log\left(\frac{\sqrt{2\pi}}en\right)+\frac5{12}\log(n)\\ &+\log\left(\frac{\sqrt{2\pi}}{A}\right)+\frac1{12}+\frac1{12n}+O\!\left(\frac{\log(n)}{n^2}\right)\tag8 \end{align} $$ and therefore, $$ \prod_{k=1}^nk!=\frac{\sqrt{2\pi}}{A}e^{1/12}\,\color{#C00}{n^{n^2/2}e^{-3n^2/4}}\color{#090}{\left(\frac{\sqrt{2\pi}}en\right)^n}n^{5/12}\color{#00F}{e^{\frac1{12n}+O\left(\frac{\log(n)}{n^2}\right)}}\tag9 $$ Finally, $$ \bbox[5px,border:2px solid #C0A000]{\left(\prod_{k=1}^nk!\right)^{1/n^2}=\color{#C00}{n^{1/2}e^{-3/4}}+\color{#090}{O\!\left(\frac{\log(n)}{n^{1/2}}\right)}}\tag{10} $$


The Glaisher–Kinkelin Constant

Equation $(6)$ is essentially Stirling's Formula: $$ \prod_{k=1}^nk=\sqrt{2\pi}\,n^{n+1/2}e^{-n}\left(1+\frac1{12n}+O\!\left(\frac1{n^2}\right)\right)\tag{11} $$ where $\sqrt{2\pi}=e^{-\zeta'(0)}$. This shows that $\zeta'(0)=-\frac12\log(2\pi)$.

Equation $(7)$ says that $$ \prod_{k=1}^nk^k=A\,n^{n^2/2+n/2+1/12}e^{-n^2/4}\left(1+O\!\left(\frac{\log(n)}{n^2}\right)\right)\tag{12} $$ where $A=e^{\frac1{12}-\zeta'(-1)}$.

Just as $$ \sum_{k=1}^n\frac1k=\log(n)+\gamma+O\!\left(\frac1n\right)\tag{13} $$ is the defining limit for $\gamma$, the Euler-Mascheroni Constant, $(12)$ appears to be the defining limit for $A$, the Glaisher–Kinkelin constant.

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Here is a more accurate than needed estimate from my answer at Formula for $1! \times 2! \times \cdots \times n!$?:

$\prod\limits_{k=0}^{n} n! \sim C^{1/2} (2\pi)^{3/8}n^{5/12}(2\pi)^{n/2}(n/e)^n \left(\dfrac{n}{e^{3/2}}\right)^{n^2/2} $ where $C =\lim\limits_{n \to \infty} \dfrac1{n^{1/12}}\prod\limits_{k=1}^n\left( \dfrac{k!}{\sqrt{2\pi k}(k/e)^k} \right) \approx 1.046335066770503 $.

Taking the $n^2$ root, $\left(\prod\limits_{k=0}^{n} n!\right)^{1/n^2} \to\left(\dfrac{n}{e^{3/2}}\right)^{1/2} $ since all the other terms go to $1$.

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  • $\begingroup$ Wow that is amazing. That is one neat solution. Love it! $\endgroup$ Commented Mar 30, 2020 at 6:26
  • $\begingroup$ Note that $C = e^{2\zeta '( - 1)} (2\pi )^{1/4}$, where $\zeta$ is Riemann's zeta function. $\endgroup$
    – Gary
    Commented Mar 30, 2020 at 9:08
  • $\begingroup$ @Gary: With your $C$, $C^{1/2}(2\pi)^{3/8}=2.1244594414121106650$, which is what I got in my answer. However, $C=1.1372658212289253252$ doesn't match the $C$ in this answer. This means that in my answer, $C_2=e^{\zeta'(-1)}\sqrt{2\pi}$. $\endgroup$
    – robjohn
    Commented Mar 30, 2020 at 15:53
  • $\begingroup$ I obtained it from the fact that the product of factorials is $ n!G(n + 1)$, where $G$ is the Barnes $G$-function. The known asymptotics of $G(n+1)$ and $n!$ yields the asymptotics of the product with the constant I gave (it is related to the Glaisher--Kinkelin constant). The ratio of my $C$ and marty cohen's $C$ is $e^{1/12}$. $\endgroup$
    – Gary
    Commented Mar 30, 2020 at 17:48
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    $\begingroup$ @Gary: I have improved my approach using another answer which approximates the zeta function using the Euler-Maclaurin Sum Formula. The constants are in terms of $\zeta'(0)$ and $\zeta'(-1)$, which are given in terms of $\pi$ and $A$. $\endgroup$
    – robjohn
    Commented Mar 31, 2020 at 7:00
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If the Riemann sum is convergent, that means that the left Riemann sum limit is the same as the right Riemann sum limit. That means $$\lim_{n\to \infty}\frac1n\sum_{r=1}^n \frac{r-1}n\ln\frac{r-1}n=\lim_{n\to \infty}\frac1n\sum_{r=1}^n \frac{r}n\ln\frac{r}n$$ Now just use the squeeze theorem, since each term in $\frac{r-1}{n}\ln\frac rn$ is between the corresponding terms in the left and right Riemann sums.

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  • $\begingroup$ Great! Just one thing what is left and right reimann sum? $\endgroup$ Commented Mar 30, 2020 at 4:40
  • $\begingroup$ Your sum formula counts the larger numbers more than the smaller ones. The factorial formula weights the smaller numbers more: $n$ shows up once, $n-1$ shows up twice, ... $2$ shows up $n-1$ times and $1$ shows up $n$ times. $\endgroup$
    – robjohn
    Commented Mar 30, 2020 at 4:40
  • $\begingroup$ @HrishabhNayal en.wikipedia.org/wiki/Riemann_sum. You draw rectangles on each small interval. The height of the rectangle is either the function on the left end or the right end. The integral must be the same in the limit $n\to\infty$ $\endgroup$
    – Andrei
    Commented Mar 30, 2020 at 5:07
  • $\begingroup$ @robjohn Correct. The full sum should have been $\frac{n-(r-1)}n\ln\frac rn$. You can split this into two sums, one for $\ln\frac rn$ and one for the sum that the question was about $\endgroup$
    – Andrei
    Commented Mar 30, 2020 at 5:15
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If we take logarithm of sequence in question we get another sequence $$a_n=\frac{1}{n^2}\left(\sum_{k=1}^{n}\log k! +xn^2\log n\right)\tag{1}$$ Applying Cesaro-Stolz we can see that limit of $a_n$ is same as that of $$\frac{1} {2n-1}\left(\log n! +xn^2\log n-x(n-1)^2\log n-x(n-1)^2\log(1-1/n)\right) $$ provided the limit of above exists. The above expression has the same limit as that of $$\frac{1}{2n-1}(\log n! +2nx\log n)+\frac{x}{2}\tag{2}$$


We can now use Stirling approximation which says that $$\log n! - \left(n\log n - n+\frac{1}{2}\log(2\pi n) \right) \to 0$$ which implies that $$\frac{1}{2n-1}(\log n! - n\log n+n) \to 0$$ Now expression $(2)$ can be rewritten as $$\frac{\log n! - n\log n+n} {2n-1}+(1+2x)\frac{n\log n}{2n-1}-\frac{n}{2n-1}+\frac {x} {2}$$ Clearly the first term tends to $0$ and the last two terms have a finite limit $(x-1)/2$. The second term has a finite limit only when $x=-1/2$. If $x<-1/2$ the expression in $(2)$ tends to $-\infty $ and if $x>-1/2$ the expression tends to $\infty$.

It follows that the limit in question exists if $x\leq - 1/2$. If $x<-1/2$ then the desired limit is $0$ and if $x=-1/2$ then the desired limit is $e^{(x-1)/2}=e^{-3/4}$.


Another alternative is to apply Cesaro-Stolz on first term in $(2)$. Doing this gives us the expression $$\frac{1}{2}\left(\log n+2xn\log n-2x(n-1)\log n-2x(n-1)\log(1-1/n)\right)$$ which has same limit as that of $$\frac{1+2x}{2}\log n+x$$ and the conclusion is same as obtained earlier. This approach avoids complicated Stirling formula.

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  • $\begingroup$ Can you explain how you used Cesaro-Stolz in second step. i. e how you wrote in form of sequence ${a_{n+1}-a_n \over b_{n+1}-b_n}$? $\endgroup$ Commented Mar 30, 2020 at 11:40
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    $\begingroup$ @HrishabhNayal, for the expression above the Stirling's approximation, you have $\log(n-1)=\log\left(n\cdot\frac{n-1}{n}\right)=\log n+\log\left(1-\frac{1}{n}\right)$, while $\log(n!)-\log((n-1)!)=\sum_{i=1}^n\log i-\sum_{i=1}^{n-1}\log i=\log n$ $\endgroup$
    – PinkyWay
    Commented Mar 30, 2020 at 12:16
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    $\begingroup$ Or just $\log\left(\frac{n!}{(n-1)!}\right)=\log n$ $\endgroup$
    – PinkyWay
    Commented Mar 30, 2020 at 12:24
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    $\begingroup$ @ ms._VerkhovtsevaKatya Thanks! I see it now $\endgroup$ Commented Mar 30, 2020 at 12:32

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